Every smooth manifold $M$ has a PL structure, and therefore a triangulation. Given a submanifold $N$ of $M$, does anyone know some nice conditions for $N$ to be the subcomplex of some triangulation of $M$, or isotopic to one?

$\begingroup$ Perhaps I am missing an important adjective, but I think your first sentence is false: arxiv.org/abs/1303.2354 $\endgroup$ – Adam Saltz May 10 '15 at 20:57

1$\begingroup$ @Adamyou missed the important adjective "smooth"; with that adjective it's a famous theorem of Whitehead. Manolescu's theorem says that there is a (highdimensional) topological manifold without a simplicial triangulation. It was already known by KirbySiebenmann that there are topological manifolds that are not PL (more restrictive than just triangulable) and by CassonTaubesFreedman that there are nontriangulable topological 4manifolds. $\endgroup$ – Danny Ruberman May 11 '15 at 0:27
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It follows from Verona's solution to Thom's triangulation conjecture that the inclusion $N\hookrightarrow M$ is triangulable whenever it is proper and topologically stable, and $M$ and $N$ are without boundary.

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$\begingroup$ @IgorRivin: That's a good question, I confess to not being an expert. Here's what I know: $\operatorname{Emb}(M,N)$ is open in $C^\infty(M,N)$, and dense if $2m\le n$. Also, the space of topologically stable maps $M\to N$ is open and dense in $C^\infty(M,N)$ (the ThomMather theorem). I believe it follows that every embedding is isotopic to a topologically stable, hence triangulable, embedding. $\endgroup$ – Mark Grant May 12 '15 at 1:13

$\begingroup$ I realize this is now a somewhat old question, but doesn't this follow even more directly from Theorem 7.8 in Verona's book by treating the pair (M,N) as giving an abstract stratification. The bundle condition is satisfied due to the tubular neighborhood theorem. Is there a reason that's not right? $\endgroup$ – Greg Friedman Oct 29 '15 at 23:12

$\begingroup$ @GregFriedman: You may be right. I don't have access to the book right now, what is the precise statement of 7.8? $\endgroup$ – Mark Grant Oct 30 '15 at 7:20

$\begingroup$ Theorem 7.8 says "Let A be an a.s [abstract stratification] of finite depth. Then there exists a smooth triangulation $(K,\phi)$ of $A$ [then there's a bit more about how you can choose it to so that a certain map to a manifold is simplicial if you also want that]." Of course it takes some unwinding of the definitions to know what a smooth triangulation is and what an abstract stratification is. But it's basically a ThomMather space, and if $N$ is a proper smooth submanifold of $M$, I think the pair should satisfy the ThomMather conditions. $\endgroup$ – Greg Friedman Oct 31 '15 at 0:41